Webslinger

Hi, Ok, I’ll have to concede that Im a newbie as far as calculating resistance and need to get straight the issue of ohms, concerning multiple speakers. This is what I think is the right info…

.1- With two speakers wired in SERIES (current running through eachother [+/- <---> +/-] ) The total resistance is multiplied by 2, therefore two 8ohm speakers are a 16ohm load. And two 16ohm speakers are a 32ohm load. Is this correct?

.2- With two speakers wired in PARALLEL (+ & - of each is separate of eachother, but both +s & -s are tied together) The total resistance is divided in half*. Therefore two 8ohn speakers are a 4ohm load and two 16s become 8. Is this correct?
(*This is a law of physics I cant understand, but thats another story)

.3 On, Marshall amps (and others) that have two 1\4” speaker output jacks, the two outputs are designed & internally wired parallel with eachother. Therefore, a combo amp, with two onboard 16ohm speakers powered from one jack, and a 2x12 extension cabinet with twin16ohm speakers powered from the other jack. Is a total 4ohm load on the amp? Is this correct?

.4- 8ohm speakers are usually/generally used in single-speaker combos, or twin-speaker combos not intended to be used with extension speakers/cabinets. True?

.5- Most 2x12 combos & 2x12 extensions, generally use 16ohm speakers in parallel, to produce a 8ohm load, because that is the most common load. True?

.6- 4x12 cabinets, I imagine can have either 8 or 16’s, wired either way?

Am I on the right track here?

bigragu63

Welcome Webslinger!

You're on the right track!

On point 1, you're close. While multiplying by 2 works when you have two speakers with the same impedance, the simple formula is that the values are simply added together when wired in series. For example: if you wired four 4 ohm speakers in series, you would have a 16 ohm load. R=R1+R2+R3...

On point 2, close again. It's simple with two speakers having the same impedance, and your example is correct. But, what if your 2X12 combo has two 16 Ohm speakers wired in parallel (therefore an 8 ohm load), and you have a 16 ohm extension cab?
The formula is R=R1xR2/R1+R2 In this example, the combined impedance of the 2 loads would be 5.33 ohms. In this case, you'd set the amp's impedance selector to 4 ohms, according to Jim. Here's a link to what Jim has to say about speaker impedance; Marshall Amps :: Frequently Asked Questions

Point 3: You are correct. Assuming the two 16 ohm speakers in the amp, and the two 16 ohm speakers in the extension cab are indeed wired in parallel, each creating a seperate 8 ohm load, then the load on the amp would be 4 ohms.

Point 4: Don't know.

Point 5: Don't know.

Point 6: You can put any speakers you want in a 4x12 cab. The real question is, can you wire them to produce an impedance load that will match an available impedance setting on your amp? In addition to series and parallel speaker cicuits, there is another option called series/parallel, which has yet another formula to calculate impedance!

Hope this helps!